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I have a question about the concept of degree of a map from differential topological point of view:

For a smooth map $f: M \to N$ between compact smooth $n$-manifolds the degree at a regular point $y \in N$ is defined by

$$deg(f,y) = \sum_{x \in f{-1}(y)} sign(det(T_x f)$$

(compare with Milnor’s definition from „TOPOLOGY FROM THE DIFFERENTIABLE VIEWPOINT“ (page 27) Therefore it counts hor often $T_xf : T_x M \to M_y N$ preserves and changes the orientation.

Now my problem: It is a well known fact that the reflection map $R: S^n \to S^n, (x_0, x_1, …, x_n) \to (x_0, x_1, …, -x_n)$ has degree $deg(R)=-1$. Take for example the easiest case $S^n = S^1$. Obviosly it can be covered by the open subsets and corresponding chart maps $$U_x:= \{ (x,y) \in S^1 \vert x > 0 \}, \varphi_{U_x}: U_x \to \mathbb{R}, (x,y) \mapsto y$$ $$V_x:= \{ (x,y) \in S^1 \vert x < 0 \}, \}, \varphi_{V_x}: U_x \to \mathbb{R}, (x,y) \mapsto y $$ $$U_y:= \{ (x,y) \in S^1 \vert y > 0 \}, \}, \varphi_{U_y}: U_x \to \mathbb{R}, (x,y) \mapsto x $$ $$V_y:= \{ (x,y) \in S^1 \vert y < 0 \} \}, \varphi_{V_y}: U_x \to \mathbb{R}, (x,y) \mapsto x $$ Here occurs following problem: Obviously $(0,-1)$ is regular wrt the reflection map $R$ and has only one preimage, namely $R^{-1}((0,-1))= (0,1)$. Since calculation of diferential $dR_{(0,1)}$ is a local problem and $(0,1) \in U_y, (0,-1) \in V_y$ it suffice to calculate $det(dR_{(0,1)})$ it only wrt the charts $\varphi_{U_y}, \varphi_{V_y}$:

But $\varphi_{V_x} \circ f \circ \varphi_{U_x}^{-1}= id_{\mathbb{R}}$ since it maps $x \to x$. Therfore $T (\varphi_{U_x} \circ f \circ \varphi_{U_x}^{-1})= T id_{\mathbb{R}}=1$ But then $deg(R, (0,-1)) = sign(det(1))=1$. Since under some nice enough conditions $def(R)= deg(R,y)$ is independend of the choice of regular point we get $deg(R)=1$, a contradiction to $deg(R)=-1$. Could anybody help me to find the error in my reasonings.

How can I show that $deg(R)=-1$ holds using Milnor's definition of degree above? I know that the is a way using only algebraic topology without working explicitely with charts. The point is what fails in the example above?

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First, I want to say that this is a great problem -- it shows that you've really tried to get a grip on the idea of degree by working through a complete example. Yay!

The difficulty you're encountering arises because all the charts must have the same orientation before you can use them to measure degree. That is to say, for any two overlapping domains $U_a, U_b$ with charts $\phi_a$ and $\phi_b$, you need for $$ \phi_b \circ \phi_a^{-1} :\Bbb R^n \to \Bbb R^n $$ to be orientation-preserving (i.e., its derivative must have positive determinant) wherever it is defined.

[I don't have my copy of TFtDV here at home to check whether Milnor says this, but I expect it's mentioned somewhere.]

If you look at your example, you'll see that $U_x$ and $U_y$, for instance, don't share an orientation: the transition function between them has negative jacobian at every point (because increasing $x$, in the first quadrant, corresponds to decreasing $y$). If you replace the coordinate function on $U_y$ by its negative, then they'll be co-oriented.

On the other hand, $U_y$ and $V_x$ are co-oriented ... but if you do the negation in the prior paragraph, you'll need to do it for $V_x$ as well to keep them co-oriented.

Once you get that all fixed up, you'll find the that degree you want to compute really is $-1$, as expected.

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  • $\begingroup$ Hi, thank you very much for this enlightening answer. So the thing that makes the machinery works correctly is to choose oriented charts? Here, for example the both stereographic projections instead of the non oriented four charts as above? $\endgroup$ – KarlPeter Dec 16 '18 at 21:03
  • $\begingroup$ You could use stereographic projection. Or you could simply redefine $\phi_{V_x} (x, y) = -y$ and $\phi_{U_y} (x, y) = -x$. BTW, in your displayed set of four things defining the $\phi$ functions, the domains of all four $\phi$ functions are written as $U_x$, but should in fact change from line to line. $\endgroup$ – John Hughes Dec 16 '18 at 22:02

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